The elves at Santa’s Workshop are trying to wrap the biggest box they can with the last bit of wrapping paper they have. All of the boxes they have are rectangular prisms with a base whose length is three times the width. Assume that they can make the box have any height they wish. There is 240 square inches of wrapping paper left. What dimensions will maximize the volume of the box given the amount of wrapping paper the elves have? Make sure to show all your work and write a sentence at the end to interpret your results.

The length of the box is 2×√30 inches long, the width of the box is 2×√(10/3) inches long, while the height of the box is √30 inches long

Step-by-step explanation:

The given parameters are;

The length of the last rectangular prism box = 3 × The width of the box

The area of the available wrapping paper = 240 in.²

Let L represent the length of the rectangular prism box, let W represent the width of the rectangular prism box and let h, represent the height of the rectangular box

Therefore, we have;

L = 3 × W

The surface area of the box, A = 2 × L × W + 2 × L × h + 2 × W × h

Whereby, L = 3 × W, we have;

A = 2 × 3 × W × W + 2 × 3 × W × h + 2 × W × h = 8·h·W + 6·W²

A = 8·h·W + 6·W² = 240

8·h·W + 6·W² = 240

8·h·W = 240 - 6·W²

h = (240 - 6·W²)/(8·W)

The volume,  V = L × W × h

∴ V = 3 × W × W × h = 3 × W² × h

V = 3 × W² × h

∴ V = 3 × W² × (240 - 6·W²)/(8·W) = 3/8 × W × (240 - 6·W²) = 90·W - 9/4·W³

dV/dW = d(90·W - 9/4·W³)/dW = 90 - 27/4·W²

At the maximum point, dV/dW = 0, which gives;

dV/dW = 90 - 27/4·W² = 0 at maximum point

27/4·W² = 90

W² = 4/27 × 90 = 40/3

W = ±√(40/3) = ±2×√(10/3)

Differentiating again and substituting gives;

d²V/dW² = d(90 - 27/4·W²)/dW =  - 27/4·W

Given that when W = 2×√(10/3), d²V/dW² is negative, and when W = -2×√(10/3), d²V/dW² is positive, W = 2×√(10/3) is the local maximum point

Therefor, the dimensions of the box are;

W = 2×√(10/3)

L = 3 × W = 3 × 2×√(10/3)

L = 3 × 2×√(10/3) = 2×√30

L = 2×√30

L = 6 ×√(10/3)

h = (240 - 6·(2×√(10/3))²)/(8·(2×√(10/3))) = √30

h = √30

Therefore, the length of the box is 2×√30 inches long, the width of the box is 2×√(10/3) inches long, while the height of the box is √30 inches long.

too hard the other guy did it

step-by-step explanation: